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where
g
n
,
s
=
α
(
n
−
s
)
1
+
s
β
.
n
−
Since the
k
-th from the last stage corresponds to the
(
m
−
k
+
1
)
-th (from the first)
stage, the player A's optimal strategy for the
k
-th stage (1
≤
k
≤
m
)is
1
r
k
α
+
n
−
1
r
k
α
+
n
−
1
a
k
−
n
,
.
+
+
r
k
g
n
,
s
r
k
g
n
,
s
n
On the other hand, the expected payoffs of player B's mixed strategy
b
=(
q
,
1
−
q
)
against player A's pure strategies are
v
k
−
1
,
m
)+(
v
k
−
1
,
m
)
E
(
Recon
,
b
)=
q
(
r
m
−
k
+
1
α
+
1
−
q
)(
1
−
n
+
v
k
−
1
,
m
)
.
E
(
¬
Recon
,
b
)=
qr
m
−
k
+
1
(
−
β
)+(
1
−
q
)(
1
+
(8.5)
Since we obtain
n
n
r
m
−
k
+
1
g
n
,
s
q
=
n
v
k
−
1
,
m
+
(
α
+
β
)+
+
r
m
−
k
+
1
n
from (
8.4
)and(
8.5
), similar to
a
k
, the player B's optimal strategy
b
k
=(
q
,
1
−
q
)
for
≤
≤
the
k
-th stage (1
k
m
)is
n
r
k
g
n
,
s
+
n
r
k
g
n
,
s
+
b
k
n
,
1
−
.
n
For simplicity, we assume
g
n
,
s
α
+
s
β
. Then, the approximation can be derived
as follows. If
k
is very large or
α
<
β
n
holds, we have
1
n
,
n
−
1
a
k
b
k
(
,
1
,
0
)
.
n
On the other hand, if
n
α
<
β
holds, we have
s
β
α
+
α
α
+
a
k
b
k
(
β
,
,
0
,
1
)
.
s
s
β
If
n
r
k
α
and
β
α
hold, we have
sn
1
2
+
1
2
n
−
1
1
+
s
a
k
b
k
n
,
,
s
,
.
(
2
+
s
)
(
2
+
s
)
n
+
2
+
s
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